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\markboth{\hfil Blow-up of radially symmetric solutions \hfil
EJDE--2002/11} {EJDE--2002/11\hfil Dimitrios E. Tzanetis \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Electronic Journal of Differential Equations}, Vol. {\bf
2002}(2002), No. 11, pp. 1--26. \newline ISSN: 1072-6691. URL:
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu (login: ftp)}
\vspace{\bigskipamount} \\
%
Blow-up of radially symmetric solutions of a non-local problem
modelling Ohmic heating
%
\thanks{ {\em Mathematics Subject Classifications:}
35B30, 35B40, 35K20, 35K55, 35K99. \hfil\break\indent
{\em Key words:} Nonlocal parabolic equations, blow-up, global existence,
steady states.
\hfil\break\indent \copyright 2002 Southwest Texas State
University. \hfil\break\indent
Submitted October 2, 2001. Published February 1, 2002.
\hfil\break\indent
Partially supported by the E.C. Human Capital and Mobility Scheme,
under contract \hfil\break\indent ERBCHRXCT 93-0409.
} }
\date{}
%
\author{ Dimitrios E. Tzanetis }
\maketitle
\begin{abstract}
We consider a non-local initial boundary-value problem
for the equation
$$ u_t=\Delta u+\lambda f(u)/\Big(\int_{\Omega}f(u)\,dx\Big)^2 ,\quad
x \in \Omega \subset \mathbb{R}^2 ,\,\;t>0,
$$
where $u$ represents a temperature and $f$ is a positive and
decreasing function. It is shown that for the radially symmetric
case, if $\int_{0}^{\infty}f(s)\,ds 0$
such that for $\lambda>\lambda^{\ast}$ there is no stationary
solution and $u$ blows up, whereas for $\lambda0\,, \label{enaa}
\\
\mathcal{B} (u):=\frac {\partial u}{\partial n}+\beta
(x)u=0\,,\quad t>0\,,\;x\in\partial\Omega\,, \label{enab}
\\
u(x,0)=u_{0} (x)\,,\quad x\in \Omega\,, \nonumber
\end{gather} \label{ena}
\end{subequations}
where $u=u(x,t)$, $\Omega$ is a bounded domain of
$\mathbb{R}^2$, $\lambda$ is a positive parameter, $\partial
\Omega $ and $\beta (x)$ are sufficiently smooth. The function $f$
is continuous, positive and decreasing,
\begin{equation}
f(s)>0\,,\quad f'(s)<0\,,\quad s\geq 0\,. \label{duo}
\end{equation}
We also study, in Section 3, the case of $f$ being the Heaviside
function (which is neither continuous nor always positive). The
equation (\ref{enaa}) arises by reducing the system of two
equations \begin{subequations}
\begin{gather}
u_t=\nabla \cdot (k(u)\nabla u)+\sigma (u)\,|\nabla \phi|^2
\label{tria}\\
\nabla \cdot(\sigma (u)\nabla \phi)=0\,,\label{trib}
\end{gather}
\label{tri} \end{subequations} to a simple, but still realistic
equation. More precisely, (\ref{tria}) is a parabolic equation
while (\ref{trib}) is an elliptic, $u$ represents the temperature
produced by an electric current flowing through a conductor,
$\phi=\phi (x,t)$ is the electric potential, $k(u)$ is the thermal
conductivity and $\sigma (u)$ is the electrical conductivity.
Problem (\ref{tri}) models many physical situations especially in
thermistors \cite{ac1,ac2,all,fow}, fuse wires, electric arcs and
fluorescent lights. The conductivity $\sigma $ may be either
decreasing or increasing in $u $ depending upon the nature of the
conductor. Here we consider materials having constant thermal
conductivity, e.g. $k(u)=1$, and decreasing electrical
conductivity, the latter allowing a thermal runaway to take place
\cite{lac9,lac95}.
Some questions concerning the steady problem to (\ref{tri}) were
investigated by Cimatti \cite{cim89,cim90,cim99}, see also
\cite{all}. A similar problem to (\ref{tri}) with radial symmetry,
Robin boundary conditions of the form $u_{n}+\beta u=0$ and
conductivity $\sigma (u)=\exp(-f(u)/\epsilon)$, $\epsilon\ll 1$
was discussed by Fowler et al. \cite{fow}. Some numerical results
are also given for small $\beta$. See also Howison \cite{how} for
how the steady problem to (\ref{tri}) may
be reduced to one nonlinear o.d.e. and Laplace's equation.\\
Carrillo \cite{car}, has looked at the bifurcation diagram of the
non-local elliptic problem with decreasing nonlinearity and
Dirichlet boundary conditions, in $\Omega \subset \mathbb{R}^N$,
See \cite{beb} for a similar study where $\Omega$ is a unit ball
in $\mathbb{R}^N $. For an extended study of the structure of
solutions of the non-local elliptic problem see \cite{lo}.
The two-dimensional mathematical problem for the
single equation can be derived by considering a long and thin
cylindrical conductor $D$, $(x,y,z)\in D \subset \mathbb{R}^3$, of
length $L$, $R\ll L$ where $R$ is the radius of the cross-section
$\Omega$ of $D$, see Figure \ref{sxh1}.
\begin{figure}
\begin{center}
\includegraphics[width=0.8\textwidth]{fig1.eps}
\end{center}
\caption{A long and thin cylindrical conductor} \label{sxh1}
\end{figure}
The curved surface of $D$ is electrically insulated, i.e. $\phi
_n=0$, and $u=0$ or more generally $u_n+\beta u=0$ for $\beta \in
\left[0,\infty\right]$, ($\beta =0$ gives $u_n=0$ while $\beta
=\infty$ gives $u=0$). Also $\phi (x,y,0,t)=0$, $\phi (x,y,L,t)=V$
at the ends of the conductor, thus $V$ is the potential
difference. Here the temperature $u$ is taken to be initially
independent of $z$ ($u=0$ is likely to be of practical interest).
Also the $z$-derivatives of $u$ are neglected, so the model gives
$z$-independence of $u$ for $t>0$. Moreover we stress that the
model is most definitely only supposed to apply in the bulk of the
device. Thus taking the thermal conductivity to be constant, and
neglecting the end effects, problem (\ref{tri}) can then be
reduced, as in \cite{lac9}, to a single non-local equation
\begin{equation}
u_t =\Delta u + \lambda \sigma (u)/(\int_ \Omega \sigma
(u)\,dx)^2, \label{tes} \end{equation} where $\lambda =I^2
/|\Omega|^2 \geq 0$, $I$ is the electric current which we suppose
to be constant and $|\Omega |$ is the measure of $\Omega $. On the
other hand, by assuming the voltage $V$ to be constant, $\phi_
x=V/L$, problem (\ref{tri}) takes the more standard semi-linear
parabolic form:
\begin{equation}
u_t =\Delta u+ \lambda \sigma
(u)\,,\;\;x\in\Omega,\;\;\mbox{where}\;\;\lambda = V^2/L^2\geq 0.
\label{pen} \end{equation} Finally, taking the more general case
of a conductor connected in series with a resistance $R_0$ under a
constant voltage $E$, then (\ref{tri}) gives, on using
\[
E=I\,R_0 +V=\Big[I+R_0\, |\Omega|\int_ \Omega \sigma (u)\,dx\Big]\,V\,,
\]
the non-local equation
\begin{equation}
u_ t =\Delta u + \lambda \sigma (u)/\Big[a+b\int_ \Omega \sigma
(u)\,dx\Big]^2 ,\,\; x \in\Omega\,,\;\,t>0,\;\, a,b>0. \label{eks}
\end{equation}
For the derivation of equations (\ref{tri})-(\ref{eks}), as well
as a complete study of the one-dimensional model for a decreasing
$\rho (u)$, (the electrical resistivity $\sigma (u)=1/\rho (u)$ is
increasing), see \cite{lac9,lac95}. In \cite{lac9,lac95} it was
shown that for $\int_{0}^{\infty} f(s)\,ds\lambda^*$ there is no steady state and $u$ blows up globally, for
$\lambda =\lambda ^{\ast}$, and $f(s)=\exp(-s)$, again there is
no steady state and $u$ exists globally in time but is unbounded.
Moreover, for $\lambda 0$ provided now that $\int_{0}^{\infty}f(s)\,ds=\infty$, where a
unique steady state exists, this steady state is globally
asymptotically stable. A global existence and divergence result
for the solution of (\ref{ept}) (see below), when $f(s)=e^{-s}$,
is also proved in \cite{kth1}.
Chafee \cite{cha} considered a related model
$u_t=u_{xx}-g(u)+\lambda f(u)/ (\int_{-1}^{1} f(u)\,dx)^2$. It was
found that there is a $\lambda^* $ such that for $\lambda
0\,,
\label{epta}\\
u(1,t)=u_r (0,t)=0\,,\quad t>0\,,\label{eptb} \\
u(r,0)=u_{0} (r)\,,\quad 00$ is
positive,
moreover if
$u_0(x)=u_0(r)$ and $\Omega $ is a ball, then $u(x,t)=u(r,t) $ is
radially symmetric and satisfies (\ref{ept}) with the proper
boundary conditions ((\ref{eptb}) or (\ref{oxto})). Furthermore if
$u_0'(r) < 0$ then $u_r (r,t)<0$, $00$, $f(s)\geq c>0$ and Lipschitz for $s\in (a,b)$, where
$a\Delta \overline u+\frac{\lambda f(\overline
u)}{(\int_\Omega f(\overline u)\,dx)^2} \,, \quad \mbox{in}
\;\Omega\,, \quad t>0\,, \\
\mathcal{B}(\overline u)>\mathcal{B}(u)=0 \quad \mbox{on} \quad
\partial
\Omega\,,\;\; t>0\,, \\
\overline u_{0} (x)>u_{0}(x)\,,\;\;\mbox{in}\;\; \Omega\,,
\end{gather}
\label{dduo} \end{subequations} while if $\underline u$ satisfies
the reversed inequalities of (\ref{dduo}) it is called a strict
\bf lower solution\rm. Now if we set $v=\overline u-\underline u$
then there exists $T>0$ such that
\begin{equation}
\begin{gathered}
v_{t}>\Delta v+\frac{\lambda f'(s)}{(\int_\Omega f(\underline
u)\,dx)^2}\,v\,,\;\;x\in
\Omega\,,\;\;00 \;\;\mbox{at}\;\;\; t=0 \;\;\mbox{in}\;\,\Omega
\;\;\mbox{and}\quad \mathcal{B}
(v)>0\,,\;\;\mbox{on}\;\;\,\partial \Omega\,,\;\;00$ at $t=T$. Moreover, if
(\ref{dduo}) holds with $\geq$, then (\ref{dtri}) also holds with
$\geq$ in the place of $>$. As long as $\underline u\,,\;\;
\overline u$ exist and $\overline u\geq \underline u$, with $f$
Lipschitz, we can apply iteration schemes similar to those of
Sattinger \cite{sat}, to show that there exists a unique solution
$u$ to (\ref{ena}) such that $\underline u\leq u\leq\overline u$.
If now $f$ is increasing then some of the above results can be
adapted by using a pair of upper-lower solutions; see \cite{lac9}.
\section{The Heaviside function}
We consider now $f(s)$ to
be the Heaviside function (decreasing), $f(s)=H(1-s)$, then
$f(s)=1$ for $s<1$, and $f(s)=0$ for $s\geq1\,,\:$ which is
neither strictly positive nor Lipschitz continuous. Thus problem
(\ref{ept}) becomes, \begin{subequations}
\begin{gather}
u_t=\Delta_r u+\lambda H(1-u)/4\pi ^2
\Big(\int_0^1H(1-u)r\,dr\Big)^2,\quad 00\,,
\label{tenaa}\\
u(1,t)=u_r (0,t)=0\,, \quad t>0\,, \quad u(r,0)=u_0 (r)\,, \quad
00,
\label{tduo} \end{equation} or there exists an or some
$s=s(t),\;\, 00\,, \\
u=1\,, \quad u_r=0, \quad 0\leq r\leq s\,,\;\; t>0\,,
\end{gather}
\label{ttri} \end{subequations} where $\pi (1-s^2 (t))=m(t)$. Note
that $u$ is continuous and $u_r \leq 0$, the latter follows by
using the maximum principle. The corresponding steady state to
(\ref{tduo}) is
\begin{equation}
\Delta_r w+\lambda /\pi ^2 =0\,, \quad 0\lambda \;\, $
and $s(t_1 )=0$, for $t_1 0\,,\\
b(0)=0\,,\quad \mbox{where}\;\; 0\leq b\leq \min\{1,\lambda /4\pi
^2 \}\,.
\end{gathered}
\label{toct} \end{equation} Then we have,
\[
\mathcal{E}(\underline v)=\dot{b}(t)(1-r^2)+4b-\frac{\lambda}{4\pi
^2}\leq \dot{b}(t)+4b- \frac{\lambda}{4\pi ^2}=0\,,
\]
on taking $\dot{b}(t)+4b-\frac{\lambda}{4\pi ^2}=0$, since
$\dot{b}(t)>0$, giving $b(t)=\frac{\lambda}{4\pi
^2}(1-e^{-4t})\to \frac{\lambda}{4\pi ^2}$ as $ t\to \infty$.
Thus for $0\lambda $
and $s\to S_0 +\,, \;\;\overline \lambda (s)\to \lambda =\lambda
(S_0)$, as $t\to \infty$.
Also we construct a lower solution $\underline v$
increasing in time, having a similar form to that as in the proof
of the blow-up (see below), in particular like (\ref{toct})
followed by a complementary version of (\ref{texi}). But now
$\dot{s}(t)>0\,,\;\; \dot{\underline \lambda}=\underline \lambda
'(s)\dot{s}>0$ for $t>t_1\,,\;\;s(t_1)=0$ and $s(t)\to S_0
-\,,\;\;\underline \lambda (s) \to \lambda = \lambda (S_0) $, as
$t\to \infty$. Hence $\underline v\leq u\leq\overline v,u$ exists
for all time and $u\to w(r;S_0)$ the unique solution for $4\pi ^2
\leq\lambda =\lambda (S_0)<8\pi ^2$, which is globally
asymptotically stable.
We show now that the solution $u$, ``blows up" (it ceases to be
less than $1$ in $[0,1)$, we recall that $u\leq1$ in $(0,1)$
as long as $u_0(r)\leq1$ ) in the sense that it becomes $1$ in
$[0,1)$ in finite time, for $\lambda
>8\pi ^2$. Therefore we get a lower solution $\underline v(r,t)$ of the
form:\hspace{.3cm} $\underline v(r,t)=b(t)(1-r^2)\,,\quad
0t_1\,,\;\;\,\mbox{and} \\
\underline v(r,t)=\frac{1+2s^2 \ln r-r^2}{1+2s^2 \ln
s-s^2}\,,\quad s(t)t_1\,,
\end{gather*}
$b(t_1)=1$, $s(t_1)=0$, $\underline v(r,t_1)=1-r^2$. Then we have,
$$\mathcal{E}(\underline v)=\left\{\begin{array}{ll}
\displaystyle 0\,, & 0\leq r\leq
s(t)\,,\; t\geq t_1\,,\\
\displaystyle \underline v_t +\frac{\underline \lambda -
\lambda}{\pi ^2 (1-s^2)^2}\leq 0\,, & s(t)\leq r\leq
1\,,\;\;t>t_1\,,
\end{array}\right.
$$
provided that $s(t)$ satisfies
\[
00$, and
$s(t)\to 1-\,,$ $\underline \lambda (s) \to 8\pi ^2 -$, as $t\to
T^{\ast}0$, $f'(s)<0$ for $s\geq
0$ and $\int_0^\infty f(s)\,ds=1$. The corresponding steady
problem to (\ref{ept}) for $f(s)=e ^{-s}$ is
\begin{equation}
w''(r)+\frac {1}{r}w'(r)+\mu e ^{-w(r)} =0\,,\quad 01$,
$M=\sup\|w\|=w(0)=2 \ln \alpha$. The parameter $\lambda$ is given
by
\begin{equation}
\lambda =4\pi ^2 \mu \Big(\int_0^1 e^{-w} r\,dr\Big)^2
=8\big(1-\frac {1}{\alpha} \big)\pi ^2=8\pi ^2
\left(1-e^{-M/2}\right)<8\pi ^2\,, \label{Ttes}
\end{equation}
so $\alpha=\frac{\lambda^* }{\lambda^* -\lambda }$, $\lambda
=\lambda (M) \to 8\pi ^2=\lambda ^{\ast}$ as $M\to \infty$, or
equivalently as $\alpha \to \infty $, and $\lambda '(M)>0$. For
each $M$ there is a corresponding unique solution $w(r)$(this
follows from a shooting argument). Finally from the above we get
the diagram of Figure \ref{sxh3}.
\begin{figure}
\begin{center}
\includegraphics[width=0.5\textwidth]{fig3.eps}
\end{center}
\caption{Response diagram for (4.1), $f(s)=\exp(-s)$.}
\label{sxh3}
\end{figure}
If $\lambda \to \lambda^{\ast} -$, which implies that $\alpha \to
\infty$, then the solution $w(r)= 2\ln [\alpha (1-r^2)+r^2)] \to
\infty$, for every compact subset of $[0,1)$. We see below that
this also holds for a general decreasing $f$.
\subsection{Stability for {\boldmath $\lambda \frac{\lambda^{\ast}}{\lambda^{\ast} -\lambda}$. Furthermore we require
$v(r,0)=2 \ln [\alpha_0 (1-r^2)+r^2] \geq u_0(r)$. It is
sufficient to choose
$$\alpha_0 =\max\{\frac{\lambda^*}{\lambda^*
-\lambda}\,,\,\sup_{r} \frac{\exp(u_0 (r)/2) -r^2}{1-r^2}\}\,.
$$
Also $\mathcal{B}(v)\geq\mathcal{B}(u)\;\mbox{on}\;\partial \Omega $,
in fact it is $v(1,t)=u(1,t)=0$. The calculations are like these
of the one-dimensional case \cite{lac9,lac95}; we find an upper
solution $v$ decreasing in time, $v\geq u$ and $v(r,t)\to 2\ln
[A(1-r^2)+r^2]=w(r;\lambda )$ as $t \to \infty\,,\; \alpha (t) \to
A =\frac{\lambda^*}{\lambda^* -\lambda}$ as $t \to \infty$ (see
(4.6)).
In a similar way we construct a lower solution $z(r,t)$ increasing in
time. Again $z(r,t)=2\ln [\alpha (t)(1-r^2)+r^2]$ is a lower
solution provided that $\alpha(t)$ satisfies (\ref{Tpen}) and
$\alpha _0 -\frac{\lambda^*}{\lambda^* -\lambda}<0$, $\alpha (t)$
is of the form of (\ref{Texi}). Also we require $z(r,0)\leq u_0
(r)$. It is sufficient to choose $\alpha_0
=\min\{\frac{\lambda^*}{\lambda^* -\lambda}\,,\inf_{r} \frac{\exp
(u_0 (r)/2)-r^2} {1-r^2}\}$. But on $\partial \Omega \;\;
z(r,t)=u(r,t)=0$, which finally implies that $z\leq u$. Hence for
$0\lambda^{\ast} $} }
To prove that the solution $u(r,t)$ blows up for $\lambda
>\lambda^* =8\pi ^2$, we construct a lower solution which blows up.
Again we take as a lower solution a function with a similar form
to the steady state $w(r):$ $z(r,t)=w(r;\mu (t))=2\ln [\alpha
(t)(1-r^2)+r^2]$. We first note that if $\alpha (t)$ satisfies
(\ref{Tpen}) and $\alpha_0 0$, moreover $z(r,t)$ is an unbounded lower
solution to (\ref{ept}) and $z(r,t) \to \infty \;\;\mbox{as}\; t
\to \infty$ for any $r\in [0,1)$. This implies that $u(r,t)$ is
unbounded, more precisely $\limsup_{t\to t^{\ast}}\|u(\cdot,t)\|
\to\infty\,,\quad t^{\ast}\leq\infty$. To prove that
$t^{\ast}0$, $0

1$ and $\alpha-1\geq \alpha /k$. The last condition is
satisfied for $t\geq t_1$, for some $t_1$ since the use of the
lower solution $z$ above guarantees unboundedness of $u$ and
allows $Z$ to be large for $t\geq t_1$. For $\lambda >\lambda^* $
and $1

0$. Furthermore from (\ref{Tdek}) we obtain,
\begin{equation}
4(\Lambda -1)(t-t_1)=\int_{\alpha (t_1)}^{\alpha
(t)}s^{p-3}\,ds \alpha /k$. The relation (\ref{Tent})
implies that $\alpha (t)\:$ blows up at
\[
T^{\ast}=t_1+\frac{\alpha_1 ^{p-2}}{4(\Lambda -1)(2-p)}0\;\;\mbox{in}\;\;0\leq r
\leq 1-\delta$. Also taking into account
(\ref{aux})-(\ref{pept}),
\begin{eqnarray}
\lefteqn{ z''(r)+\frac{1}{r}z'+\mu f(z)}\nonumber\\
&=&\frac{z'}{r}+\mu(f(z)-g(z))>\frac{z'(r)} {1-\delta}+\mu
(f(z)-g(z))
\label{pokt}\\
&=&-\frac{\sqrt{2\mu}\;[G(z)-G(M)]^{1/2}}{1-\delta}+\mu
(f(z)-g(z))\,,\;\;\mbox{in} \;\;1-\delta\mu (f(z)-g(z))-\mu (f(z)-g(z))=0\,.
\]
In addition $z'(0)=z(1)=w'(0)=w(1)=0$, hence $z$ is a lower
solution to $w$-problem. This implies
\begin{equation}
z(r)\leq w(r)\;\;\mbox{and}\;\; w'(1)\leq z'(1)<0\,, \label{pdek}
\end{equation} (if the latter inequality were $w'(1)>z'(1)$ it would give
$z(r)>w(r)$ for some $r,\,$ which would be a contradiction).\\
Now taking:
\begin{enumerate}
\item[(a)] $g$, such that
$01-2\epsilon$, $\epsilon >0$, from the definition of
$G$,
\item[(c)] $\mu$ to satisfy (\ref{pene}).
\end{enumerate}
Note that $G'(z)=-g(z) <0$, $G(z)$ is decreasing and $G(0)\leq
1$); from (\ref{ptri}), (\ref{pexi}) and (\ref{pdek}) we obtain
\[
2>\frac{(w'(1))^2}{\mu }\geq
\frac{(z'(1))^2}{\mu}=2[G(0)-G(M)]>2(1-2\epsilon )\,.
\]
This relation holds for every $\epsilon >0$, as far as $\mu \gg
1$, hence this proves the lemma.\quad\hfill$\Box$\medskip
\begin{prop}\label{PrDUO}
If (\ref{ptes}) holds then $\,\lambda 0\,, \label{pdod} \end{equation} then
problem (\ref{pena}) has a unique solution.
\end{prop}
\paragraph{Proof:}
From (\ref{pduo}) we get
$\lambda (\mu)=4\pi ^2 \frac{(w'(1))^2}{\mu}=4\pi ^2 (W'
(1))^2$ by writing $w(r)=\nu W(r)$, $\nu =\sqrt \mu$. Then
(\ref{pena}) gives $W'_{\nu}(0)=W_{\nu}(1)=W'(0)=W(1)=0$ and
\[
W''+\frac{W'}{r}+\nu f(w)=0\quad \mbox{or}\quad
W''_{\nu}+\frac{W'_{\nu}}{r}+\nu ^2
f'(w)W_{\nu}=-f(w)-f'(w)w\,.\quad
\]
If (\ref{pdod}) holds, then using maximum principle and Hopf's
boundary lemma, we get that $W_{\nu} (r)>0$ and $W'_{\nu}(1)<0$
or $\frac{d}{d\nu} (W'(1))^2 >0 $, since also $W'(1)<0\,.$ Then
$\lambda '(\mu)=\frac{2\pi ^2}{\nu}\frac{d}{d\nu}(W'(1))^2
>0\;\; $ which implies uniqueness.\quad\hfill$\Box$ \medskip
\paragraph{Remark:} Proposition \ref{PrTES} is also true for a
general domain $\Omega$. The relation (\ref{pdod}) implies
(\ref{pent}).
Finally we obtain the response diagram of Figure \ref{sxh4}.
\begin{figure}
\begin{center}
\includegraphics[width=0.5\textwidth]{fig4.eps}
\end{center}
\caption{Response diagram for the Dirichlet problem,
$\int_0^\infty f(s)\,ds=\infty$.}\label{sxh4}
\end{figure}
\subsection {Stability where a unique steady state exists}
We follow the same procedure as in the previous section, we seek
for a decreasing (increasing)-in-time upper (lower) solution to
problem (\ref{ept}). We first look for an upper solution of a form
similar to the steady state: $v(r,t)=w(r;\overline
{\mu}(t))=\overline w$, where $\overline w$ is a steady state.
Then
\begin{equation}
\mathcal{E}(v)=\overline w_{\overline \mu}\:\dot{\overline
\mu}+[4\pi ^2 \overline \mu \:I^2 (\overline w)-\lambda ]
f(\overline w) /4\pi ^2 I^2 (\overline w)\;,\smallskip
\label{pdtr}
\end{equation}
where $I(w)=\int_0^1 f(w)r\,dr$ and $-\Delta _r w=\mu f(w)$
($\Delta_r =\partial ^2/\partial r^2 +\frac{1}{r}\partial
/\partial r$). Also $\overline w_{\overline \mu}=\frac{\partial
\overline w}{\partial \overline \mu }>0$, $w(r;\overline \mu (t))$
is increasing with respect to $\overline \mu$, by using the
maximum principle. Taking now any $\lambda>0 $, so that
$w(x;\lambda)$ is the unique stationary solution, we can choose
$\overline \mu (0)$ such that $w(r;\overline \mu (0))=v(r,0)\geq
u_0 (r)\,;$ this can be done since $u_0 (r),\; u'_0 (r)\:$ are
bounded. Furthermore, since a unique steady state exists (see
Proposition \ref{PrTES} ) for these values of $\lambda$, there
exists a $\mu$ such that $\lambda =4\pi ^2 \mu I^2 (w)$,
$M=w(0;\mu )=w(0)$, where $w(r)= w(r;\mu )$ is the unique steady
state of problem (\ref{pena}), and as long as $\overline \mu >\mu$
then $\overline w>w$ and $\overline \lambda=\overline \lambda (t)
=4\pi ^2 \overline \mu (t) I^2 (\overline w)>\lambda$. As before
$w(r;\overline \mu)$ is an upper solution, decreasing in time,
which tends to the stationary solution $w$ provided that
\begin{equation}
0\mu $ (note that $f(s)$ is bounded away
from zero and $w_{\mu}$ is also finite). Hence $ u=u(r,t)\leq
v(r,t)=w(r;\overline \mu (t))$ and $\dot{\overline \mu}<0$ which
implies that $v_t =\overline w_{\overline \mu}\:\dot{\overline
\mu} <0$. In a similar way we can construct a lower solution
$z(r,t)=w(r;\underline \mu (t))$ which is increasing in time and
tends to the steady state $w$. Finally we obtain, $
z(r,t)=w(r;\underline \mu (t))\leq u(r,t)\leq v(r,t)=
w(r;\overline \mu (t))$, and $\overline \mu (t)\to \mu
+\,,\;\underline \mu (t) \to \mu-\,,\; v(r,t)\to w+\,, \; z(r,t)
\to w-$, as $t\to\infty$. This implies that $u(\cdot ,t) \to
w(\cdot )$ uniformly as $t\to \infty$, and that $w$ is globally
asymptotically stable.\smallskip\\ For the case of $\lambda \geq
\lambda^*$ and $\int_0^\infty f(s)\,ds=1$, there is no steady
solution to (\ref{pena}), then $\underline \lambda =\underline
\lambda(t) =4\pi ^2 \underline \mu(t) I^2 (\underline w) 0$. Taking the above lower solution $z(r,t)$ we obtain that
$\underline \mu (t)\to \infty $ $(\dot{\underline \mu}>0)$ and $
u(r,t)\geq z(r,t)=w(r;\underline \mu(t)) \to \infty$ as $t\to
\infty$ (note that $w_{\mu} >0$ and $\mu (t) \to \infty$ as
$t\to \infty$ then $z \to \infty$ as $t\to \infty)$. In
particular $\sup_{r}u(r,t)\to \infty$ as $t\to t^*\leq\infty$,
hence $u$ is unbounded.
\subsection{Blow-up for {\boldmath $\lambda >\lambda^* $}}
We can now prove that the solution $u$ to problem (\ref{ept})
blows up in finite time if $\lambda >\lambda^* =8\pi ^2. $ To
prove this we use similar methods to those in the previous
sections, (or see \cite{kth,lac9, lac95}). We look for a lower
solution $z(r,t)$ to the $u$-problem which itself blows up. We try
to find a lower solution with a similar form to the steady state.
We take into account the form of blow-up in the one-dimensional
case, therefore we consider $z(r,t)$ to satisfy:
\begin{subequations}
\begin{gather}
z(r,t)=M(t)=\sup_{r}z(r,t)\,,\quad z_r (r,t)=0\,,\quad 0\leq r\leq
1-\delta (t)\, ,\;\;t>0\,, \label{depena}\\
z_{rr}+\mu (t) g(z)=0\;,\quad1-\delta (t)\leq r\leq 1\,,\;\;z(1,t)
=0\,,\;\; t>0\,, \label{depenb}
\end{gather} \label{depen}\end{subequations}
where $01/[(\lambda /8\pi ^2)^{1/2}-1]$ for
$\lambda>\lambda^* =8\pi ^2 $. Such an $\alpha$ gives $\Lambda
=\frac{1}{3}[\lambda /\pi ^2 (1+\alpha)^2 -\frac{8}{\alpha
^2}]>0$.
\begin{lemma}\label{PPEN}
$ M\,f(M)\to 0 $ as $M\to \infty$.
\end{lemma}
For the proof of this lemma, see \cite{lac95}.
\begin{lemma}\label{PEXI}
$\delta \to 0$ as $ \mu \to \infty$.
\end{lemma}
\paragraph{Proof:}
From the previous lemma we have that $f(M)\to 0$ and $g(M)\to 0$
as $M\to \infty$ and for fixed $\alpha$, (\ref{pden}) implies
that $\mu \to \infty$ as $M\to \infty\,.$ For $00\,,\;\;\beta >0\,.
\label{sunor}
\end{equation}
The corresponding steady problem is
\begin{subequations}\begin{gather}
w''+\frac{1}{r}w'+\mu f(w)=0\,,\quad 00\;,\quad w'(0)=0\,.
\end{gather}
\label{exis} \end{subequations} Multiplying again by $w'$ and
integrating we obtain
\begin{equation}
\frac{(w'(1))^2}{2}+\int_0^1 \frac{(w'(r))^2}{r}\,dr-\mu \int_m^M
f(s)\,ds=0\,, \label{sunth} \end{equation} where $0\lambda^*$. Thus we have the diagrams of
Figure \ref{sxh5}.
\begin{figure}
\begin{center}
\includegraphics[width=0.9\textwidth]{fig5.eps}
\end{center}
\caption{Possible non-local response diagrams for the Robin
problem.}\label{sxh5}
\end{figure}
\subsection{Stability} We consider, as in the Dirichlet
problem, an upper solution $v(r,t)=w(r;\overline \mu (t))$ which
is decreasing in time and a lower solution $z(r,t)= w(r;\underline
\mu (t))$ which is increasing in time, to the $u$-problem. More
precisely we have $\mathcal{E}(v)\geq 0$, and $\mathcal{E}(z)\leq
0$ provided that
\begin{gather*}
0u_0(r)$, $w(r;\underline
\mu (0))0$, if $\mu_1\mu_2$ and $\Phi(\mu
,\lambda )=0$ if $\mu=\mu_1$ or $\mu =\mu_2 $.
For $\lambda =\lambda^*$, $\Phi (\underline \mu ,\lambda )<0$,
hence $z_1(r,t)\to w^ {\ast}$ as $t\to \infty$, provided
that $w_1(r;\underline \mu (0))w^{\ast}$. This means that $w^{\ast}$ is unstable. More
precisely it is unstable from above and stable from below. For
$\lambda>\lambda^* $ again $\Phi(\underline \mu ,\lambda )<0$,
$z(r,t) \to \infty$ as $t\to t^{\ast}\leq \infty$, hence $u$ is
unbounded for any initial data; this also holds even for
$\lambda\lambda^* $ or for
$\lambda\leq \lambda^*$ but with initial conditions larger than
the greatest steady state. Following the same method as in the
one-dimensional case \cite{lac95}, if $u$ fails to blow-up, then
for any given $k$, there must be a $t_k>0$ such that $u\geq k$,
for $t_k\geq t$ (this is due to the use of the lower solutions,
note that $m=\min w(r)=w(1)\to \infty$ as $M\to \infty )$. Then
we consider the problem,
\begin{gather*}
v_t=\Delta _r v+\lambda f_k (v)/4\pi ^2 (\int_0^1 f_k
(v)r\,dr)^2\,,\;\;00 \\
v(1,t)=v_r (0,t)=0\,,\;\;t\geq t_k \\
v(r,t_k)=0\,,\;\;0\lambda^* \int_0^\infty f_k
(s)\,ds= \lambda^* \int_0^\infty f(s+k)\,ds=\lambda^*
\int_k^\infty f(S)\,dS$, then $v$ blows up at finite time. Hence
choosing $k$ sufficiently large so that the previous inequality
holds, we get that $u$ blows up. This blow-up is global.
Finally, carrying over the analysis similar to the
Dirichlet problem and the one-dimensional case \cite{lac9,lac95},
we obtain Figure \ref{sxh6}. We use the notation: $(\to \cdot
\leftarrow )$, for stable stationary solutions, $(\leftarrow
\cdot \to )$ for unstable, and the double arrows $(\to \to )$
for solutions $u$ which blow up. If we lie in the region where
$\Phi (\overline \mu (t),t)>0$ then the arrows point downwards
while where $\Phi (\underline \mu (t),t)<0$ the arrows point
upwards.
\begin{figure}
\begin{center}
\includegraphics[width=\textwidth]{fig6.eps}
\end{center}
\caption{Stability and blow-up of solutions for the Robin
problem.}\label{sxh6}
\end{figure}
Any solution which corresponds to a point of the curves of this
type $ (\to \cdot \leftarrow)$ is stable while all others are
unstable. More precisely this $(\to \cdot \to)$, is stable from
one side and unstable from the other whereas this $(\leftarrow
\cdot \to)$, is unstable from both sides.
For the Neumann problem (the boundary conditions
are $u_r (0,t)=u_r (1,t)=0)$ there is no positive steady state
for any $\lambda>0$. Concerning the solution $u(r,t)$, this
behaves as in the one-dimensional case \cite{lac9,lac95}; if
$\int_0^\infty f(s)\,ds0$, $f'(u)<0$, $u$, represents the temperature which
is produced in a conductor having fixed electric current $I$,
i.e. $\lambda =I^2 /\pi ^2$. The function $f$ represents
electrical conductivity $(f(u)=\sigma (u))$, in contrast to the
one-dimensional model where it represents electrical resistivity
$(f(u)=\rho (u)=1/\sigma (u))$. This work extends the results of
the one-dimensional problem, and the methods used are similar to
the one-dimensional case and are based on comparison techniques.
We find similar behaviour in both problems, and it is rather like that of
the standard reaction-diffusion model, $u_t=\Delta u+\lambda
f(u)$, $f(u)>0$, $f'(u)>0$, $f''(u)>0$, see \cite{lac} and the
references therein. More precisely, for the Dirichlet problem, if
$\int_0^\infty f(s)\,ds \lambda^* $ there is no
steady state and $u$, blows up globally. Also in case we have a
unique steady solution, this solution is asymptotically stable.
For the Robin problem, provided that $\int_0^\infty
f(s)\,ds\lambda^*$.
Concerning the stability, the minimal solution is stable, the next
greater one is unstable, the next stable and so on. On the other
hand, if $\lambda>\lambda^* $ then $u$ blows up; $u$ also blows
up for $\lambda \in (0,\lambda^* )$ and for sufficiently large
initial data . The solution(s) at $\lambda=\lambda^* $ is(are)
unstable.\\ For the Neumann problem there is no steady state and
the solution $u$ blows up in finite time if $\int_0^\infty
f(s)\,ds